Sequences and Series – Exercise 9.1 – Class XI

  1. Write the first terms of the sequences whose nth terms is an = n(n + 2)

Solution:

an = n(n + 2)

Substituting  n = 1, 2, 3, 4 and 5 we obtain

a1 = 1(1+2) = 3

a2 = 2(2+2) = 8

a3 = 3(3+2) = 15

a4 = 4(4+2) = 24

a5 = 5(5+2) = 35

Therefore, the required terms are 3, 8, 15, 24 and 35.


  1. Write the first five terms of the sequences whose nth terms is an = n/(n+1)

Solution:

an = n/(n+1)

Substituting n = 1, 2, 3, 4, 5 obtain

a1 = 1/1+1 = 1/2 ,

a2 = 2/2+1 = 2/3 ,

a3 = 3/3+1 = 3/4 ,

a4 = 4/4+1 = 4/5 ,

a5 = 5/5+1 = 5/6 ,

Therefore, the required terms are 1/2, 2/3 , 3/4 , 4/5  and 5/6


  1. Write the first five terms of the sequences whose nth term is an = 2n

Solution:

an = 2n

Substituting n = 1, 2, 3, 4, 5 we obtain

a1 = 21 = 2

a2 = 22 = 4

a3 = 23 = 8

a4 = 24 = 16

a5 = 25 = 32

Therefore the required terms are 2, 4, 8, 16 and 32


  1. Write the first five terms of the sequences whose nth terms is an = (2n-3)/6

Solution:

Substituting n = 1, 2, 3 , 4, 5 we obtain

a1 = (2×1-3)/6 = -1/6 ,

a2 = (2×2-3)/6 = 1/6 ,

a3 = (2×3-3)/6 = 3/6 = 1/2 ,

a4 = (2×4-3)/6 = 5/6 ,

a5 = (2×5-3)/6 = 7/6 ,

Therefore, the required terms are -1/6 , 1/6 , 1/2 , 5/6  and 7/6 .


  1. Write the first five terms of the sequences whose nth terms is an = (-1)n-15n+1

Solution:

Substituting n = 1, 2, 3, 4, 5 we obtain

a1 = (-1)1-1 51+1 = 52 = 25

a2 = (-1)2-1 52+1 = -53 = -125

a3 = (-1)3-1 53+1 = 54 = 625

a4 = (-1)4-1 54+1 = -55 = -3125

a5 = (-1)5-1 55+1 = 56 = 15625

Therefore, the required terms are 25, -125, 625, -3125 and 15625.


  1. Write the first five terms of the sequences whose nth term is an = n[(n2+5)/4]

Solution:

Substituting n = 1, 2, 3, 4, 5 we obtain

a1 = 1. [(12+5)/4] = 6/4 = 3/2,

a2 = 2. [(22+5)/4] = 2. 9/4 = 9/2,

a3 = 3. [(32+5)/4] = 3. 14/4 = 21/2,

a4 = 4. [(42+5)/4] = 21,

a5 = 5. [(52+5)/4] = 5. 30/4 = 75/2,

Therefore, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.


  1. Find the 17th in the following sequences whose nth term is an = 4n – 3 , a17, a24

Solution:

Substituting n = 17 we obtain

a17 = 4(17) – 3 = 68 – 3 = 65

Substituting n = 24 we obtain

a24 = 4(24) – 3 = 96 – 3 = 93


  1. Find the 7th term in the following sequences whose nth terms is an = [(n2)/2n]; a7

Solution:

Substituting n = 7 we obtain

a7 = (72)/(27) = 49/128


  1. Find the 9th term in the following sequences whose nth term is an = (-1)n-1n3; a9

Solution:

Substituting n = 9 we obtain

a9 = (-1)9-1(9)3 = (9)3 = 729


  1. Find the 20th term in the following sequences whose nth terms is an = [{n(n+2)}/(n+3)]; a20

Solution:

Substituting n = 20, we obtain

a20 = [{20(20+2)}/(20+3)] = 20(18)/23 = 360/23


  1. Write the first five terms of the following sequences and obtain the corresponding series:

a1 = 3, an = 3an-1 + 2, for all n > 1

Solution:

a1 = 3, an = 3an-1 + 2 for all n>1

a2 = 3a1 + 2 = 3(3) + 2 = 11

a3 = 3a2 + 2 = 3(11) + 2 = 35

a4 = 3a3 + 2 = 3(35) + 2 = 107

a5 = 3a4 + 2 = 3(107) + 2 = 323

Hence the first five terms of the sequences are 3, 11, 35, 107 and 323.

The corresponding series 3 + 11+ 35 + 107 + 323 + …


  1. Write the first five terms of the following sequence and obtain the corresponding series:

a1 = -1, an = [(an-1)/n], n ≥ 2

Solution:

a1 = -1, an = [(an-1)/n], n ≥ 2

a2 = (a1)/2 = -1/2,

a3 = (a2)/3 = -1/6,

a4 = (a3)/4 = -1/24,

a5 = (a4)/4 = -1/120,

Hence the first five terms of the sequences are -1, -1/2, -1/6, -1/24, -1/120,

The corresponding series is (-1)+( -1/2)+( -1/6)+( -1/24)+( -1/120)+…


  1. Write the first five terms of the following sequence and obtain the corresponding series:

a1 = a2 = 2, an = an-1 -1, n > 2

Solution:

a1 = a2 = 2, an = an-1 -1, n > 2

a3 = a2 – 1 = 2 – 1 = 1

a4 = a3 – 1 = 1 – 1 = 0

a5 = a4 – 1 = 0 – 1 = -1

Hence, the first five terms of the sequence are 2, 2, 1, 0 and -1 and the corresponding series is 2 + 2 + 1 + 0 + (-1) + …


  1. The Fibonacci sequence is defined by 1 = a1= a2 and an = an-1 + an-2, n>2

Find [(an+1)/(an)], for n = 1, 2, 3, 4, 5

Solution:

1 = a1= a2

an = an-1 + an-2, n>2

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

For n = 1, [(an+1)/(an)] = [(a2)/(a1)] = 1/1 = 1

For n = 2, [(an+1)/(an)] = [(a3)/(a2)] = 2/1 = 2

For n = 3, [(an+1)/(an)] = [(a4)/(a3)] = 3/2

For n = 4, [(an+1)/(an)] = [(a5)/(a4)] = 5/3

For n = 5, [(an+1)/(an)] = [(a6)/(a5)] = 8/5


 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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